Question

A deer walking at 3.29 m/s speeds up to 13.38 m/s, covering a distance of 63.5 meters during this acceleration. How much time did it take him to do this?

Answer 1

Assuming the deer's acceleration was constant, we have

`\Delta x=\frac{v_f+v_0}2t`

where
`\Delta x` is the deer's total displacement,
`v_f,v_0` are the deer's final and initial velocities, and
`t` is time. Then

`63.5\,\mathrm m=\frac{13.38\,(\mathrm m)/(\mathrm s)+3.29\,(\mathrm m)/(\mathrm s)}2t`

`\implies t=7.62\,\mathrm s`